Answer:
(a) Mn = M₁ + (n-1) (M₂ -M₁) = 1 + (n- 1) 1 = n (b) n > 10 (exceed 10) or n =11 (c) n >50 or n= 51
After making a journey of 51 times, the rocket will be discarded
Explanation:
Solution
(a) Let Mn denotes the number of maintenance visits after the nth journey
Then M₁ = 1 , M₂ = 1 +M₁ = 2, M₃ = 1 +M₂ = 3
We therefore, notice that M follows an arithmetic sequence
So,
Mn = M₁ + (n-1) (M₂ -M₁)
= 1 + (n- 1) 1 = n
or Mn =n
(b) For what value of n we will get fro Mn > 10
Thus,
n > 10 (exceed 10) or n =11
(c)Similarly of Mn is greater than 50 or Mn>50, the rocket will not be used or reused
So,
n >50 or n= 51
After making a journey of 51 times, the rocket will be discarded
Answer:
-25.63°C.
Explanation:
We know that throttling is a constant enthalpy process
From steal table
We know that if we know only one property in side the dome then we will find the other property by using steam property table.
Temperature at saturation pressure 1 bar is 99.63°C and Temperature at saturation pressure 0.35 bar is about 74°C .
So from above we can say that change in temperature is -25.63°C.
But there is no any option for that .
Answer:
True
Explanation:
By definition of steady flow we have
where f(x,y,z,t) is any property of the system under consideration
=> f(x,y,z,t) = constant
Answer:
class TriangleNumbers
{
public static void main (String[] args)
{
for (int number = 1; number <= 10; ++number) {
int sum = 1;
System.out.print("1");
for (int summed = 2; summed <= number; ++summed) {
sum += summed;
System.out.print(" + " + Integer.toString(summed));
}
System.out.print(" = " + Integer.toString(sum) + '\n');
}
}
}
Explanation:
We need to run the code for each of the 10 lines. Each time we sum numbers from 1 to n. We start with 1, then add numbers from 2 to n (and print the operation). At the end, we always print the equals sign, the sum and a newline character.
Answer:
pls put a question not random letters
Explanation:
